boat

[Gone Angling]

I'm planning on building a cajun sneak boat. I want to use it for a fishing
situation in a small shallow bay. What is the optimal colour to paint the
submerged portion so it doesn't scare the fish ? It's a nice starter project.

[Bob La Londe]

"Gone Angling" > wrote in message
...
> I'm planning on building a cajun sneak boat. I want to use it for a
fishing
> situation in a small shallow bay. What is the optimal colour to paint the
> submerged portion so it doesn't scare the fish ? It's a nice starter
project.
>
>

How about taking a swim and looking up from under water.


--
Bob La Londe
Yuma, Az
http://www.YumaBassMan.com
Promote Your Fishing, Boating, or Guide Site for Free
Simply add it to our index page.
No reciprocal link required. (Requested, but not required)

[Chuck Coger]

Paint it a bright color like Blaze Orange or Hot Pink. The portion you are
referring to is typically called a "hull".
This isn't a joke either, rumor has it bright colored boat bottoms and side
help keep bass from jumping. Which if your in a tournament is a good thing.
I was told that Pro's enjoyed fishing out of the Kellogs boat because the
bass didn't jump as much.

Regardless of what color you paint it on somedays it will stand out on some
days as weather, water clarity, and depth of the fish all have an impact on
what the fish see's. An old timer who owned a lot of boats told me that he
caught more fish out of a red boat than any other color. He fished the same
couple of lakes for years and used that as his gauge, luckily I like red;)

---
Chuck Coger
http://www.fishin-pro.com


"Gone Angling" > wrote in message
...
> I'm planning on building a cajun sneak boat. I want to use it for a
fishing
> situation in a small shallow bay. What is the optimal colour to paint the
> submerged portion so it doesn't scare the fish ? It's a nice starter
project.
>
>
>

[Fritz Nordengren]

This is a good topic and I'm glad you brought it up.

When you consider the light refraction coefficient of the gradients, the
first thought would be a blue grey, but clearly once you run the
numbers, you see quickly why that is wrong.

In the CRC Handbook of Chemistry & Physics, page E-224 in my 56th
Edition, I
find the following formulae:

(n-1) *10^7 = const1 + const2 / lambda^2 + const3 / lambda^4

where n is the required refractive index
lambda is the wavelength in micron (millionths of a metre -- in the actual
printed formula there are additional factors of 10^8 & 10^16 to cope with
the fact that visible spectral wavelengths are usually quoted in Aangstroem)

The values of the three constants in dry air at a pressure of 1 atm are

const1 const2 const3
30 deg C 2589.72 12.259 0.2576
15 deg C 2726.43 12.288 0.3555
0 deg C 2875.66 13.412 0.3777

A correction is given in the source to allow for air which has a water
vapour content.

To a first approximation, the variation of the refractive index with
temperature and pressure can be attributed to changes in gas density, and
the gas density can be obtained from the ideal gas equation:

(n-1) proportional to m/V = pM/RT.

But that is just a first approximation, that does not hold up for more
precise work.

Atmospheric refraction slightly increases the observed elevation angle
of a peak relative to the observer. The effect is actually quite
complicated, since it depends on the precise atmospheric conditions,
including atmospheric pressure, temperature, and water vapor content,
and thus varies with time and the altitudes of the observer and the
observed peak. Fortunately, the effect of refraction is less than ~15%
of the effect due to the curvature of the Earth, and typically only
increases the observed elevation angle by less than 0.1°.

Refraction is caused by two effects. First, light likes to travel on
the path that gets to the observer in the minimum time. (Light is,
after all, the fastest thing in the Universe, so you wouldn't expect it
would like to take a longer path than it had to, right?) The speed of
light is the speed of light in a vacuum divided by the index of
refraction. Second, the index of refraction of the atmosphere depends
on atmospheric pressure and amount of water present, which change with
height in the atmosphere. Therefore light actually travels on a curved
path in the atmosphere from one object to another. The path goes higher
than the straight-line distance in order to take advantage of the faster
speed higher in the atmosphere. Because the path is so curved, the
observer must always look a bit higher to see the light rays coming back
down from that higher elevation.

Clearly refraction must depend on some power of the distance. If you
are observing something close by, light can't get to you any quicker by
travelling very far upward. However, if you are far away from an
object, light can take advantage of the faster speed at higher elevation
and deviate more significantly from a straight line.

Astraightforward calculation gives the following formula for the angular
change with distance due to refraction between the observer at elevation
Zo (measured in km) who also is observing a peak at elevation Zo:

theta = [ 1.6 * c / { 2*(1+a) } ] * d

where d is the distance in miles, theta is in radians, and

a = 2.9e-4 * exp(-Zo/10 km) / (1 + 2.9 * To / 760)

b = 2.9 * alpha / {760 * (1 + 2.9 * To / 760) }

c = a * (b - 1/10 km)

alpha = 6.5° C. / km

The calculation assumes an atmosphere whose pressure, temperature (To in
Celsius) and water content only varies with elevation, and an atmosphere
whose temperature varies with elevation with the slope -alpha. (Alpha
is defined as the rate of temperature drop with altitude, and so is
positive in the above formulae in the normal case where the temperature
drops with altitude.) "exp" is the exponential function.

Note that this calculation assumes quite a bit. The real atmosphere can
vary markedly horizontally, can have temperature inversions, can change
its humidity, and have additional components like dust that change the
index of refraction. The observer and observed peak are not always at
the same elevation assumed in the derivation of this formula. Hence
there are no guarantees that this formula will always give accurate
results. However, on average, this formula probably gives the correct
average answer. The results of this formula at sea level are given in
surveying books as the proper term to use.

The book Elementary Surveying gives the equivalent formula in terms of
the "elevation loss" in feet of the observed object with distance:

elevation loss = 0.574 * d^2,

which is said to apply to near horizontal shots. What the book doesn't
say is that this formula is only correct near sea level.

The elevation loss formula consists of two terms:
The curvature of the Earth term, with coefficient 0.662 ( = 5280 / (2 *
R) = 5280 / (2*3986) = 0.662). The 5280 converts the final units in the
above formula to feet.
The atmospheric refraction term, which is therefore taken to have a
coefficient of 0.574 - 0.662 = -0.088.

The formula above gives a coefficient of 0.088 at an elevation of 0' and
a temperature of 65° F. However, the coefficient varies markedly with
temperature and elevation.

If you've read this far, then you also know that the density of the
water has a great deal to do with the light angle of refraction.
However, the density of the author of this question is beyond any
possible scientific measurement.


Gone Angling wrote:
> I'm planning on building a cajun sneak boat. I want to use it for a fishing
> situation in a small shallow bay. What is the optimal colour to paint the
> submerged portion so it doesn't scare the fish ? It's a nice starter project.
>
>

[Jerry \NervisRek\ Barton]

Whew !!!!!
--
Jerry Barton
www.jerrys-world.com

"Fritz Nordengren" > wrote in message
news:ChPqb.139951$e01.468128@attbi_s02...
> This is a good topic and I'm glad you brought it up.
>
> When you consider the light refraction coefficient of the gradients,
the
> first thought would be a blue grey, but clearly once you run the
> numbers, you see quickly why that is wrong.
>
> In the CRC Handbook of Chemistry & Physics, page E-224 in my 56th
> Edition, I
> find the following formulae:
>
> (n-1) *10^7 = const1 + const2 / lambda^2 + const3 / lambda^4
>
> where n is the required refractive index
> lambda is the wavelength in micron (millionths of a metre -- in the
actual
> printed formula there are additional factors of 10^8 & 10^16 to cope
with
> the fact that visible spectral wavelengths are usually quoted in
Aangstroem)
>
> The values of the three constants in dry air at a pressure of 1 atm
are
>
> const1 const2 const3
> 30 deg C 2589.72 12.259 0.2576
> 15 deg C 2726.43 12.288 0.3555
> 0 deg C 2875.66 13.412 0.3777
>
> A correction is given in the source to allow for air which has a
water
> vapour content.
>
> To a first approximation, the variation of the refractive index with
> temperature and pressure can be attributed to changes in gas
density, and
> the gas density can be obtained from the ideal gas equation:
>
> (n-1) proportional to m/V = pM/RT.
>
> But that is just a first approximation, that does not hold up for
more
> precise work.
>
> Atmospheric refraction slightly increases the observed elevation
angle
> of a peak relative to the observer. The effect is actually quite
> complicated, since it depends on the precise atmospheric conditions,
> including atmospheric pressure, temperature, and water vapor
content,
> and thus varies with time and the altitudes of the observer and the
> observed peak. Fortunately, the effect of refraction is less than
~15%
> of the effect due to the curvature of the Earth, and typically only
> increases the observed elevation angle by less than 0.1°.
>
> Refraction is caused by two effects. First, light likes to travel
on
> the path that gets to the observer in the minimum time. (Light is,
> after all, the fastest thing in the Universe, so you wouldn't expect
it
> would like to take a longer path than it had to, right?) The speed
of
> light is the speed of light in a vacuum divided by the index of
> refraction. Second, the index of refraction of the atmosphere
depends
> on atmospheric pressure and amount of water present, which change
with
> height in the atmosphere. Therefore light actually travels on a
curved
> path in the atmosphere from one object to another. The path goes
higher
> than the straight-line distance in order to take advantage of the
faster
> speed higher in the atmosphere. Because the path is so curved, the
> observer must always look a bit higher to see the light rays coming
back
> down from that higher elevation.
>
> Clearly refraction must depend on some power of the distance. If
you
> are observing something close by, light can't get to you any quicker
by
> travelling very far upward. However, if you are far away from an
> object, light can take advantage of the faster speed at higher
elevation
> and deviate more significantly from a straight line.
>
> Astraightforward calculation gives the following formula for the
angular
> change with distance due to refraction between the observer at
elevation
> Zo (measured in km) who also is observing a peak at elevation Zo:
>
> theta = [ 1.6 * c / { 2*(1+a) } ] * d
>
> where d is the distance in miles, theta is in radians, and
>
> a = 2.9e-4 * exp(-Zo/10 km) / (1 + 2.9 * To / 760)
>
> b = 2.9 * alpha / {760 * (1 + 2.9 * To / 760) }
>
> c = a * (b - 1/10 km)
>
> alpha = 6.5° C. / km
>
> The calculation assumes an atmosphere whose pressure, temperature
(To in
> Celsius) and water content only varies with elevation, and an
atmosphere
> whose temperature varies with elevation with the slope -alpha.
(Alpha
> is defined as the rate of temperature drop with altitude, and so is
> positive in the above formulae in the normal case where the
temperature
> drops with altitude.) "exp" is the exponential function.
>
> Note that this calculation assumes quite a bit. The real atmosphere
can
> vary markedly horizontally, can have temperature inversions, can
change
> its humidity, and have additional components like dust that change
the
> index of refraction. The observer and observed peak are not always
at
> the same elevation assumed in the derivation of this formula. Hence
> there are no guarantees that this formula will always give accurate
> results. However, on average, this formula probably gives the
correct
> average answer. The results of this formula at sea level are given
in
> surveying books as the proper term to use.
>
> The book Elementary Surveying gives the equivalent formula in terms
of
> the "elevation loss" in feet of the observed object with distance:
>
> elevation loss = 0.574 * d^2,
>
> which is said to apply to near horizontal shots. What the book
doesn't
> say is that this formula is only correct near sea level.
>
> The elevation loss formula consists of two terms:
> The curvature of the Earth term, with coefficient 0.662 ( = 5280 /
(2 *
> R) = 5280 / (2*3986) = 0.662). The 5280 converts the final units in
the
> above formula to feet.
> The atmospheric refraction term, which is therefore taken to have a
> coefficient of 0.574 - 0.662 = -0.088.
>
> The formula above gives a coefficient of 0.088 at an elevation of 0'
and
> a temperature of 65° F. However, the coefficient varies markedly
with
> temperature and elevation.
>
> If you've read this far, then you also know that the density of the
> water has a great deal to do with the light angle of refraction.
> However, the density of the author of this question is beyond any
> possible scientific measurement.
>
>
> Gone Angling wrote:
> > I'm planning on building a cajun sneak boat. I want to use it for
a fishing
> > situation in a small shallow bay. What is the optimal colour to
paint the
> > submerged portion so it doesn't scare the fish ? It's a nice
starter project.
> >
> >
>

[J Buck]

<In the CRC Handbook of Chemistry & Physics, page E-224 in my 56th
Edition, I
find the following formulae:
(n-1) *10^7 =3D const1 + const2 / lambda^2 + const3 / lambda^4
where n is the required refractive index lambda is the wavelength in
micron ......snip.... =A0 =A0 =A0 =A0 =A0 =A0 =A0 2726.43 =A0 =A0 =A0
=A0 12.288 =A0 =A0 =A0 =A0 0.3555
0 deg C =A0 =A0 =A0 =A0 2875.66 =A0 =A0 =A0 =A0 13.412 =A0 =A0 =A0 =A0
0.3777>

Huh? Lmao

[kenneth blevins]

LMAO
"J Buck" > wrote in message
...
<In the CRC Handbook of Chemistry & Physics, page E-224 in my 56th
Edition, I
find the following formulae:
(n-1) *10^7 = const1 + const2 / lambda^2 + const3 / lambda^4
where n is the required refractive index lambda is the wavelength in
micron ......snip.... 2726.43
12.288 0.3555
0 deg C 2875.66 13.412
0.3777>

Huh? Lmao

[Ken Fortenberry]

Fritz Nordengren wrote:

> This is a good topic and I'm glad you brought it up.
>
> When you consider the light refraction coefficient of the gradients, the
> first thought would be a blue grey, but clearly once you run the
> numbers, you see quickly why that is wrong. <analysis snipped>

So it would be sky blue then ? ;-)

--
Ken Fortenberry

[Steve Erwin]

Aint sure what I just read but it sounds like the motto: If ya can't dazzle
em with brilliance, baffle em with bull****. :) I might as well been trying
to read Aramic. :) :)

"Fritz Nordengren" > wrote in message
news:ChPqb.139951$e01.468128@attbi_s02...
> This is a good topic and I'm glad you brought it up.
>
> When you consider the light refraction coefficient of the gradients, the
> first thought would be a blue grey, but clearly once you run the
> numbers, you see quickly why that is wrong.
>
> In the CRC Handbook of Chemistry & Physics, page E-224 in my 56th
> Edition, I
> find the following formulae:
>
> (n-1) *10^7 = const1 + const2 / lambda^2 + const3 / lambda^4
>
> where n is the required refractive index
> lambda is the wavelength in micron (millionths of a metre -- in the actual
> printed formula there are additional factors of 10^8 & 10^16 to cope with
> the fact that visible spectral wavelengths are usually quoted in
Aangstroem)
>
> The values of the three constants in dry air at a pressure of 1 atm are
>
> const1 const2 const3
> 30 deg C 2589.72 12.259 0.2576
> 15 deg C 2726.43 12.288 0.3555
> 0 deg C 2875.66 13.412 0.3777
>
> A correction is given in the source to allow for air which has a water
> vapour content.
>
> To a first approximation, the variation of the refractive index with
> temperature and pressure can be attributed to changes in gas density, and
> the gas density can be obtained from the ideal gas equation:
>
> (n-1) proportional to m/V = pM/RT.
>
> But that is just a first approximation, that does not hold up for more
> precise work.
>
> Atmospheric refraction slightly increases the observed elevation angle
> of a peak relative to the observer. The effect is actually quite
> complicated, since it depends on the precise atmospheric conditions,
> including atmospheric pressure, temperature, and water vapor content,
> and thus varies with time and the altitudes of the observer and the
> observed peak. Fortunately, the effect of refraction is less than ~15%
> of the effect due to the curvature of the Earth, and typically only
> increases the observed elevation angle by less than 0.1°.
>
> Refraction is caused by two effects. First, light likes to travel on
> the path that gets to the observer in the minimum time. (Light is,
> after all, the fastest thing in the Universe, so you wouldn't expect it
> would like to take a longer path than it had to, right?) The speed of
> light is the speed of light in a vacuum divided by the index of
> refraction. Second, the index of refraction of the atmosphere depends
> on atmospheric pressure and amount of water present, which change with
> height in the atmosphere. Therefore light actually travels on a curved
> path in the atmosphere from one object to another. The path goes higher
> than the straight-line distance in order to take advantage of the faster
> speed higher in the atmosphere. Because the path is so curved, the
> observer must always look a bit higher to see the light rays coming back
> down from that higher elevation.
>
> Clearly refraction must depend on some power of the distance. If you
> are observing something close by, light can't get to you any quicker by
> travelling very far upward. However, if you are far away from an
> object, light can take advantage of the faster speed at higher elevation
> and deviate more significantly from a straight line.
>
> Astraightforward calculation gives the following formula for the angular
> change with distance due to refraction between the observer at elevation
> Zo (measured in km) who also is observing a peak at elevation Zo:
>
> theta = [ 1.6 * c / { 2*(1+a) } ] * d
>
> where d is the distance in miles, theta is in radians, and
>
> a = 2.9e-4 * exp(-Zo/10 km) / (1 + 2.9 * To / 760)
>
> b = 2.9 * alpha / {760 * (1 + 2.9 * To / 760) }
>
> c = a * (b - 1/10 km)
>
> alpha = 6.5° C. / km
>
> The calculation assumes an atmosphere whose pressure, temperature (To in
> Celsius) and water content only varies with elevation, and an atmosphere
> whose temperature varies with elevation with the slope -alpha. (Alpha
> is defined as the rate of temperature drop with altitude, and so is
> positive in the above formulae in the normal case where the temperature
> drops with altitude.) "exp" is the exponential function.
>
> Note that this calculation assumes quite a bit. The real atmosphere can
> vary markedly horizontally, can have temperature inversions, can change
> its humidity, and have additional components like dust that change the
> index of refraction. The observer and observed peak are not always at
> the same elevation assumed in the derivation of this formula. Hence
> there are no guarantees that this formula will always give accurate
> results. However, on average, this formula probably gives the correct
> average answer. The results of this formula at sea level are given in
> surveying books as the proper term to use.
>
> The book Elementary Surveying gives the equivalent formula in terms of
> the "elevation loss" in feet of the observed object with distance:
>
> elevation loss = 0.574 * d^2,
>
> which is said to apply to near horizontal shots. What the book doesn't
> say is that this formula is only correct near sea level.
>
> The elevation loss formula consists of two terms:
> The curvature of the Earth term, with coefficient 0.662 ( = 5280 / (2 *
> R) = 5280 / (2*3986) = 0.662). The 5280 converts the final units in the
> above formula to feet.
> The atmospheric refraction term, which is therefore taken to have a
> coefficient of 0.574 - 0.662 = -0.088.
>
> The formula above gives a coefficient of 0.088 at an elevation of 0' and
> a temperature of 65° F. However, the coefficient varies markedly with
> temperature and elevation.
>
> If you've read this far, then you also know that the density of the
> water has a great deal to do with the light angle of refraction.
> However, the density of the author of this question is beyond any
> possible scientific measurement.
>
>
> Gone Angling wrote:
> > I'm planning on building a cajun sneak boat. I want to use it for a
fishing
> > situation in a small shallow bay. What is the optimal colour to paint
the
> > submerged portion so it doesn't scare the fish ? It's a nice starter
project.
> >
> >
>

[Illinois Fisherman]

Melt fluorocarbon line in a glue gun and smear it all over da boat. Fish
can't see ya commin dat way. Optimum water refectory color dats what
fluorocarbon can do for ya.


"Gone Angling" > wrote in message
...
> I'm planning on building a cajun sneak boat. I want to use it for a
fishing
> situation in a small shallow bay. What is the optimal colour to paint the
> submerged portion so it doesn't scare the fish ? It's a nice starter
project.
>
>

[G. M. Zimmermann]

>I'm planning on building a cajun sneak boat. I want to use it for a fishing
>situation in a small shallow bay. What is the optimal colour to paint the
>submerged portion so it doesn't scare the fish ? It's a nice starter project.
>
>
I have a bit of a "mad scientist" streak myself and I always wondered why
the hull would have to be a single color? Ever see WWI warships with a
"Razzel Dazzel" paint scheme? Wouldn't a similar color scheme of large
irregular shapes in white, gray, blue, and black break up the outline of the
hull, disguising its size, shape, speed and direction? Of course I suspect
most of the times anything viewing a boat from below whould be seeing it as a
silouette without any color at all...

-Zimmy

[Brad Coovert]

>If you've read this far, then you also know that the density of the
>water has a great deal to do with the light angle of refraction.
>However, the density of the author of this question is beyond any
>possible scientific measurement.

LOL! Love it!
Brad Coovert, 2003 Tournament Director, Greenfield Bassmasters
Please visit our sponsors:
http://www.geocities.com/greenfieldbass/WelcomeToOurSponsorPage.htm

[Brad Coovert]

>Wouldn't a similar color scheme of large
>irregular shapes in white, gray, blue, and black break up the outline of the
>hull, disguising its size, shape, speed and direction?

If he did that, he'd just lose the boat.
Brad Coovert, 2003 Tournament Director, Greenfield Bassmasters
Please visit our sponsors:
http://www.geocities.com/greenfieldbass/WelcomeToOurSponsorPage.htm

[Gone Angling]

I think i'll go with experience rather than Fritz's impressive conjectures. I
will research what a Cajun would paint his boat. I suspect the result will be a
varying shades of green. The colour would also give you you a stealth presence
against the shorelines.

[RichZ]

All you needed to do was scroll to the last paragraph


RichZ©
www.richz.com/fishing

[Doc \(The Tin Boat King\)]

A Cajun, PAINT a boat? Try any left over house paint (color not important) or no paint at all. Kind of like your brain. Bill P.
================================================== ===============
"Gone Angling" > wrote in message ...
I think i'll go with experience rather than Fritz's impressive conjectures. I
will research what a Cajun would paint his boat. I suspect the result will be a
varying shades of green. The colour would also give you you a stealth presence
against the shorelines.

[Steve Erwin]

I saw that. I was just joking about the first12" of stuff. I aint smart
enough to know if it was jibberish or true, but smart enough to know he was
screwing with him, LOL.

"RichZ" > wrote in message
...
> All you needed to do was scroll to the last paragraph
>
>
> RichZ©
> www.richz.com/fishing
>

[Steve Erwin]

That's the best suggestion yet! Maybe his lads wouldn't see him, LOL.

"Illinois Fisherman" > wrote in message
y.com...
>
> Melt fluorocarbon line in a glue gun and smear it all over da boat. Fish
> can't see ya commin dat way. Optimum water refectory color dats what
> fluorocarbon can do for ya.
>
>
> "Gone Angling" > wrote in message
> ...
> > I'm planning on building a cajun sneak boat. I want to use it for a
> fishing
> > situation in a small shallow bay. What is the optimal colour to paint
the
> > submerged portion so it doesn't scare the fish ? It's a nice starter
> project.
> >
> >
>
>

[Steve @ OutdoorFrontiers]

Well heck, now that you explain it in simple terms like that, it makes a lot of sense........Heck,
any idiot can understand that!
--
Steve
OutdoorFrontiers
http://www.outdoorfrontiers.com
G & S Guide Service and Custom Rods
http://www.herefishyfishy.com

[Gone Angling]

Those that mock probably couldn't even drive a nail home without making it
crooked.

[Dave Norton]

Wrong again Putz! Plonk here you go again.

--
Dave Norton
Millennium Rods
"Gone Angling" > wrote in message
...
> Those that mock probably couldn't even drive a nail home without making it
> crooked.
>
>

[Illinois Fisherman]

And those that ask stupid questions and post nonsense shall be called TROLL.



"Gone Angling" > wrote in message
...
> Those that mock probably couldn't even drive a nail home without making it
> crooked.
>
>

[Jerry \NervisRek\ Barton]

Come on guys, more and more of you are breaking your vow to quit
responding to this "IDIOT".
--
Jerry Barton
www.jerrys-world.com

"Illinois Fisherman" > wrote in message
y.com...
>
> And those that ask stupid questions and post nonsense shall be
called TROLL.
>
>
>
> "Gone Angling" > wrote in message
> ...
> > Those that mock probably couldn't even drive a nail home without
making it
> > crooked.
> >
> >
>
>

[Gone Angling]

The plans call for a swivel seat. I don't deal a lot (in fact never) with
marine suppliers. Where would i get a reasonably priced swivel seat.
As for colour i've decided on a green hull and a camo deck.
The boat is so small it will probably only take a trolling motor. There should
also be on decks lights and all required safety equipment since marine cops
patrol the waters where i plan to use it.

[James Luning]

I would go to Bass pro marine catalog. It's a lot cheaper than Marine
supply.
Don't get the clamp style. Thay just won't stay put. Get the mounting plate
with brass bearing and the companion post that mounts to the seat. Be
careful to get the right height. Long hours on the boat demand the proper
height.
Consider an ajustable telescoping post too.
JL

"Gone Angling" > wrote in message
...
> The plans call for a swivel seat. I don't deal a lot (in fact never) with
> marine suppliers. Where would i get a reasonably priced swivel seat.
> As for colour i've decided on a green hull and a camo deck.
> The boat is so small it will probably only take a trolling motor. There
should
> also be on decks lights and all required safety equipment since marine
cops
> patrol the waters where i plan to use it.
>
>

[Bob Rickard]

Fritz, I would contribute my usual definitive, leading edge contributions to
this thread, but I can no longer allow myself to get involved where you
offer only gross generalizations with no attempt toward supplying pertinent
and applicable details of any kind.

--
Bob Rickard
www.secretweaponlures.com
--------------------------<=x O')))><


"Fritz Nordengren" > wrote in message
news:ChPqb.139951$e01.468128@attbi_s02...
> This is a good topic and I'm glad you brought it up.
>
> When you consider the light refraction coefficient of the gradients, the
> first thought would be a blue grey, but clearly once you run the
> numbers, you see quickly why that is wrong.
>
> In the CRC Handbook of Chemistry & Physics, page E-224 in my 56th
> Edition, I
> find the following formulae:
>
> (n-1) *10^7 = const1 + const2 / lambda^2 + const3 / lambda^4
>
> where n is the required refractive index
> lambda is the wavelength in micron (millionths of a metre -- in the actual
> printed formula there are additional factors of 10^8 & 10^16 to cope with
> the fact that visible spectral wavelengths are usually quoted in
Aangstroem)
>
> The values of the three constants in dry air at a pressure of 1 atm are
>
> const1 const2 const3
> 30 deg C 2589.72 12.259 0.2576
> 15 deg C 2726.43 12.288 0.3555
> 0 deg C 2875.66 13.412 0.3777
>
> A correction is given in the source to allow for air which has a water
> vapour content.
>
> To a first approximation, the variation of the refractive index with
> temperature and pressure can be attributed to changes in gas density, and
> the gas density can be obtained from the ideal gas equation:
>
> (n-1) proportional to m/V = pM/RT.
>
> But that is just a first approximation, that does not hold up for more
> precise work.
>
> Atmospheric refraction slightly increases the observed elevation angle
> of a peak relative to the observer. The effect is actually quite
> complicated, since it depends on the precise atmospheric conditions,
> including atmospheric pressure, temperature, and water vapor content,
> and thus varies with time and the altitudes of the observer and the
> observed peak. Fortunately, the effect of refraction is less than ~15%
> of the effect due to the curvature of the Earth, and typically only
> increases the observed elevation angle by less than 0.1°.
>
> Refraction is caused by two effects. First, light likes to travel on
> the path that gets to the observer in the minimum time. (Light is,
> after all, the fastest thing in the Universe, so you wouldn't expect it
> would like to take a longer path than it had to, right?) The speed of
> light is the speed of light in a vacuum divided by the index of
> refraction. Second, the index of refraction of the atmosphere depends
> on atmospheric pressure and amount of water present, which change with
> height in the atmosphere. Therefore light actually travels on a curved
> path in the atmosphere from one object to another. The path goes higher
> than the straight-line distance in order to take advantage of the faster
> speed higher in the atmosphere. Because the path is so curved, the
> observer must always look a bit higher to see the light rays coming back
> down from that higher elevation.
>
> Clearly refraction must depend on some power of the distance. If you
> are observing something close by, light can't get to you any quicker by
> travelling very far upward. However, if you are far away from an
> object, light can take advantage of the faster speed at higher elevation
> and deviate more significantly from a straight line.
>
> Astraightforward calculation gives the following formula for the angular
> change with distance due to refraction between the observer at elevation
> Zo (measured in km) who also is observing a peak at elevation Zo:
>
> theta = [ 1.6 * c / { 2*(1+a) } ] * d
>
> where d is the distance in miles, theta is in radians, and
>
> a = 2.9e-4 * exp(-Zo/10 km) / (1 + 2.9 * To / 760)
>
> b = 2.9 * alpha / {760 * (1 + 2.9 * To / 760) }
>
> c = a * (b - 1/10 km)
>
> alpha = 6.5° C. / km
>
> The calculation assumes an atmosphere whose pressure, temperature (To in
> Celsius) and water content only varies with elevation, and an atmosphere
> whose temperature varies with elevation with the slope -alpha. (Alpha
> is defined as the rate of temperature drop with altitude, and so is
> positive in the above formulae in the normal case where the temperature
> drops with altitude.) "exp" is the exponential function.
>
> Note that this calculation assumes quite a bit. The real atmosphere can
> vary markedly horizontally, can have temperature inversions, can change
> its humidity, and have additional components like dust that change the
> index of refraction. The observer and observed peak are not always at
> the same elevation assumed in the derivation of this formula. Hence
> there are no guarantees that this formula will always give accurate
> results. However, on average, this formula probably gives the correct
> average answer. The results of this formula at sea level are given in
> surveying books as the proper term to use.
>
> The book Elementary Surveying gives the equivalent formula in terms of
> the "elevation loss" in feet of the observed object with distance:
>
> elevation loss = 0.574 * d^2,
>
> which is said to apply to near horizontal shots. What the book doesn't
> say is that this formula is only correct near sea level.
>
> The elevation loss formula consists of two terms:
> The curvature of the Earth term, with coefficient 0.662 ( = 5280 / (2 *
> R) = 5280 / (2*3986) = 0.662). The 5280 converts the final units in the
> above formula to feet.
> The atmospheric refraction term, which is therefore taken to have a
> coefficient of 0.574 - 0.662 = -0.088.
>
> The formula above gives a coefficient of 0.088 at an elevation of 0' and
> a temperature of 65° F. However, the coefficient varies markedly with
> temperature and elevation.
>
> If you've read this far, then you also know that the density of the
> water has a great deal to do with the light angle of refraction.
> However, the density of the author of this question is beyond any
> possible scientific measurement.
>
>
> Gone Angling wrote:
> > I'm planning on building a cajun sneak boat. I want to use it for a
fishing
> > situation in a small shallow bay. What is the optimal colour to paint
the
> > submerged portion so it doesn't scare the fish ? It's a nice starter
project.
> >
> >
>

[Jerry \NervisRek\ Barton]

ROFLMAO !!!!!!!!! Old coot, you're too much!
--
Jerry Barton
www.jerrys-world.com

"Bob Rickard" > wrote in message
igy.com...
> Fritz, I would contribute my usual definitive, leading edge
contributions to
> this thread, but I can no longer allow myself to get involved where
you
> offer only gross generalizations with no attempt toward supplying
pertinent
> and applicable details of any kind.
>
> --
> Bob Rickard
> www.secretweaponlures.com
> --------------------------<=x O')))><
>
>
> "Fritz Nordengren" > wrote in message
> news:ChPqb.139951$e01.468128@attbi_s02...
> > This is a good topic and I'm glad you brought it up.
> >
> > When you consider the light refraction coefficient of the
gradients, the
> > first thought would be a blue grey, but clearly once you run the
> > numbers, you see quickly why that is wrong.
> >
> > In the CRC Handbook of Chemistry & Physics, page E-224 in my 56th
> > Edition, I
> > find the following formulae:
> >
> > (n-1) *10^7 = const1 + const2 / lambda^2 + const3 / lambda^4
> >
> > where n is the required refractive index
> > lambda is the wavelength in micron (millionths of a metre -- in
the actual
> > printed formula there are additional factors of 10^8 & 10^16 to
cope with
> > the fact that visible spectral wavelengths are usually quoted in
> Aangstroem)
> >
> > The values of the three constants in dry air at a pressure of 1
atm are
> >
> > const1 const2 const3
> > 30 deg C 2589.72 12.259 0.2576
> > 15 deg C 2726.43 12.288 0.3555
> > 0 deg C 2875.66 13.412 0.3777
> >
> > A correction is given in the source to allow for air which has a
water
> > vapour content.
> >
> > To a first approximation, the variation of the refractive index
with
> > temperature and pressure can be attributed to changes in gas
density, and
> > the gas density can be obtained from the ideal gas equation:
> >
> > (n-1) proportional to m/V = pM/RT.
> >
> > But that is just a first approximation, that does not hold up for
more
> > precise work.
> >
> > Atmospheric refraction slightly increases the observed elevation
angle
> > of a peak relative to the observer. The effect is actually quite
> > complicated, since it depends on the precise atmospheric
conditions,
> > including atmospheric pressure, temperature, and water vapor
content,
> > and thus varies with time and the altitudes of the observer and
the
> > observed peak. Fortunately, the effect of refraction is less than
~15%
> > of the effect due to the curvature of the Earth, and typically
only
> > increases the observed elevation angle by less than 0.1°.
> >
> > Refraction is caused by two effects. First, light likes to travel
on
> > the path that gets to the observer in the minimum time. (Light
is,
> > after all, the fastest thing in the Universe, so you wouldn't
expect it
> > would like to take a longer path than it had to, right?) The
speed of
> > light is the speed of light in a vacuum divided by the index of
> > refraction. Second, the index of refraction of the atmosphere
depends
> > on atmospheric pressure and amount of water present, which change
with
> > height in the atmosphere. Therefore light actually travels on a
curved
> > path in the atmosphere from one object to another. The path goes
higher
> > than the straight-line distance in order to take advantage of the
faster
> > speed higher in the atmosphere. Because the path is so curved,
the
> > observer must always look a bit higher to see the light rays
coming back
> > down from that higher elevation.
> >
> > Clearly refraction must depend on some power of the distance. If
you
> > are observing something close by, light can't get to you any
quicker by
> > travelling very far upward. However, if you are far away from an
> > object, light can take advantage of the faster speed at higher
elevation
> > and deviate more significantly from a straight line.
> >
> > Astraightforward calculation gives the following formula for the
angular
> > change with distance due to refraction between the observer at
elevation
> > Zo (measured in km) who also is observing a peak at elevation Zo:
> >
> > theta = [ 1.6 * c / { 2*(1+a) } ] * d
> >
> > where d is the distance in miles, theta is in radians, and
> >
> > a = 2.9e-4 * exp(-Zo/10 km) / (1 + 2.9 * To / 760)
> >
> > b = 2.9 * alpha / {760 * (1 + 2.9 * To / 760) }
> >
> > c = a * (b - 1/10 km)
> >
> > alpha = 6.5° C. / km
> >
> > The calculation assumes an atmosphere whose pressure, temperature
(To in
> > Celsius) and water content only varies with elevation, and an
atmosphere
> > whose temperature varies with elevation with the slope -alpha.
(Alpha
> > is defined as the rate of temperature drop with altitude, and so
is
> > positive in the above formulae in the normal case where the
temperature
> > drops with altitude.) "exp" is the exponential function.
> >
> > Note that this calculation assumes quite a bit. The real
atmosphere can
> > vary markedly horizontally, can have temperature inversions, can
change
> > its humidity, and have additional components like dust that change
the
> > index of refraction. The observer and observed peak are not
always at
> > the same elevation assumed in the derivation of this formula.
Hence
> > there are no guarantees that this formula will always give
accurate
> > results. However, on average, this formula probably gives the
correct
> > average answer. The results of this formula at sea level are
given in
> > surveying books as the proper term to use.
> >
> > The book Elementary Surveying gives the equivalent formula in
terms of
> > the "elevation loss" in feet of the observed object with distance:
> >
> > elevation loss = 0.574 * d^2,
> >
> > which is said to apply to near horizontal shots. What the book
doesn't
> > say is that this formula is only correct near sea level.
> >
> > The elevation loss formula consists of two terms:
> > The curvature of the Earth term, with coefficient 0.662 ( = 5280 /
(2 *
> > R) = 5280 / (2*3986) = 0.662). The 5280 converts the final units
in the
> > above formula to feet.
> > The atmospheric refraction term, which is therefore taken to have
a
> > coefficient of 0.574 - 0.662 = -0.088.
> >
> > The formula above gives a coefficient of 0.088 at an elevation of
0' and
> > a temperature of 65° F. However, the coefficient varies markedly
with
> > temperature and elevation.
> >
> > If you've read this far, then you also know that the density of
the
> > water has a great deal to do with the light angle of refraction.
> > However, the density of the author of this question is beyond any
> > possible scientific measurement.
> >
> >
> > Gone Angling wrote:
> > > I'm planning on building a cajun sneak boat. I want to use it
for a
> fishing
> > > situation in a small shallow bay. What is the optimal colour to
paint
> the
> > > submerged portion so it doesn't scare the fish ? It's a nice
starter
> project.
> > >
> > >
> >
>
>

[Chuck Coger]

Try Wal-Mart, they are in the fishing section that you claim to frequent.

---
Chuck Coger
http://www.fishin-pro.com


"Gone Angling" > wrote in message
...
> The plans call for a swivel seat. I don't deal a lot (in fact never) with
> marine suppliers. Where would i get a reasonably priced swivel seat.
> As for colour i've decided on a green hull and a camo deck.
> The boat is so small it will probably only take a trolling motor. There
should
> also be on decks lights and all required safety equipment since marine
cops
> patrol the waters where i plan to use it.
>
>
>

[Indian]

O.k. Fritz the world is round!! Now just how is this poor fella to break it
down into everyday language as to why he painted his "sneak boat" pink????
The Lurkin Indian.
"Fritz Nordengren" > wrote in message
news:ChPqb.139951$e01.468128@attbi_s02...
> This is a good topic and I'm glad you brought it up.
>
> When you consider the light refraction coefficient of the gradients, the
> first thought would be a blue grey, but clearly once you run the
> numbers, you see quickly why that is wrong.
>
> In the CRC Handbook of Chemistry & Physics, page E-224 in my 56th
> Edition, I
> find the following formulae:
>
> (n-1) *10^7 = const1 + const2 / lambda^2 + const3 / lambda^4
>
> where n is the required refractive index
> lambda is the wavelength in micron (millionths of a metre -- in the actual
> printed formula there are additional factors of 10^8 & 10^16 to cope with
> the fact that visible spectral wavelengths are usually quoted in
Aangstroem)
>
> The values of the three constants in dry air at a pressure of 1 atm are
>
> const1 const2 const3
> 30 deg C 2589.72 12.259 0.2576
> 15 deg C 2726.43 12.288 0.3555
> 0 deg C 2875.66 13.412 0.3777
>
> A correction is given in the source to allow for air which has a water
> vapour content.
>
> To a first approximation, the variation of the refractive index with
> temperature and pressure can be attributed to changes in gas density, and
> the gas density can be obtained from the ideal gas equation:
>
> (n-1) proportional to m/V = pM/RT.
>
> But that is just a first approximation, that does not hold up for more
> precise work.
>
> Atmospheric refraction slightly increases the observed elevation angle
> of a peak relative to the observer. The effect is actually quite
> complicated, since it depends on the precise atmospheric conditions,
> including atmospheric pressure, temperature, and water vapor content,
> and thus varies with time and the altitudes of the observer and the
> observed peak. Fortunately, the effect of refraction is less than ~15%
> of the effect due to the curvature of the Earth, and typically only
> increases the observed elevation angle by less than 0.1°.
>
> Refraction is caused by two effects. First, light likes to travel on
> the path that gets to the observer in the minimum time. (Light is,
> after all, the fastest thing in the Universe, so you wouldn't expect it
> would like to take a longer path than it had to, right?) The speed of
> light is the speed of light in a vacuum divided by the index of
> refraction. Second, the index of refraction of the atmosphere depends
> on atmospheric pressure and amount of water present, which change with
> height in the atmosphere. Therefore light actually travels on a curved
> path in the atmosphere from one object to another. The path goes higher
> than the straight-line distance in order to take advantage of the faster
> speed higher in the atmosphere. Because the path is so curved, the
> observer must always look a bit higher to see the light rays coming back
> down from that higher elevation.
>
> Clearly refraction must depend on some power of the distance. If you
> are observing something close by, light can't get to you any quicker by
> travelling very far upward. However, if you are far away from an
> object, light can take advantage of the faster speed at higher elevation
> and deviate more significantly from a straight line.
>
> Astraightforward calculation gives the following formula for the angular
> change with distance due to refraction between the observer at elevation
> Zo (measured in km) who also is observing a peak at elevation Zo:
>
> theta = [ 1.6 * c / { 2*(1+a) } ] * d
>
> where d is the distance in miles, theta is in radians, and
>
> a = 2.9e-4 * exp(-Zo/10 km) / (1 + 2.9 * To / 760)
>
> b = 2.9 * alpha / {760 * (1 + 2.9 * To / 760) }
>
> c = a * (b - 1/10 km)
>
> alpha = 6.5° C. / km
>
> The calculation assumes an atmosphere whose pressure, temperature (To in
> Celsius) and water content only varies with elevation, and an atmosphere
> whose temperature varies with elevation with the slope -alpha. (Alpha
> is defined as the rate of temperature drop with altitude, and so is
> positive in the above formulae in the normal case where the temperature
> drops with altitude.) "exp" is the exponential function.
>
> Note that this calculation assumes quite a bit. The real atmosphere can
> vary markedly horizontally, can have temperature inversions, can change
> its humidity, and have additional components like dust that change the
> index of refraction. The observer and observed peak are not always at
> the same elevation assumed in the derivation of this formula. Hence
> there are no guarantees that this formula will always give accurate
> results. However, on average, this formula probably gives the correct
> average answer. The results of this formula at sea level are given in
> surveying books as the proper term to use.
>
> The book Elementary Surveying gives the equivalent formula in terms of
> the "elevation loss" in feet of the observed object with distance:
>
> elevation loss = 0.574 * d^2,
>
> which is said to apply to near horizontal shots. What the book doesn't
> say is that this formula is only correct near sea level.
>
> The elevation loss formula consists of two terms:
> The curvature of the Earth term, with coefficient 0.662 ( = 5280 / (2 *
> R) = 5280 / (2*3986) = 0.662). The 5280 converts the final units in the
> above formula to feet.
> The atmospheric refraction term, which is therefore taken to have a
> coefficient of 0.574 - 0.662 = -0.088.
>
> The formula above gives a coefficient of 0.088 at an elevation of 0' and
> a temperature of 65° F. However, the coefficient varies markedly with
> temperature and elevation.
>
> If you've read this far, then you also know that the density of the
> water has a great deal to do with the light angle of refraction.
> However, the density of the author of this question is beyond any
> possible scientific measurement.
>
>
> Gone Angling wrote:
> > I'm planning on building a cajun sneak boat. I want to use it for a
fishing
> > situation in a small shallow bay. What is the optimal colour to paint
the
> > submerged portion so it doesn't scare the fish ? It's a nice starter
project.
> >
> >
>

[Fritz Nordengren]

Now Rickard, I've been polite to you on this group to gave you a second
chance, but surely you remember that incident in the late 80's at Los
Alamos where the entire physics staff discredited your so called
"theories" on string cheese as a fossil fuel.

Rickard, when will you learn? there are those of us who are real
scientists and not some bass fishing hobbiest who thinks that just
because he knows E=Mc2 can tell those of us in the know how to
differentiate a calculated mass.

And also, don't forget, who was there to save your butt in Las Vegas
during the 1991 International Conference on Refractive Indulgence. If
Ihadn't spent three hours correcting your formulas, you would have been
driven from the scientific community for life. If you didn't have those
pictures of me with those three girls and the sheep, I never would have
covered for you.

So lets just let it pass and agree to disagree.

<g>



Bob Rickard wrote:
> Fritz, I would contribute my usual definitive, leading edge contributions to
> this thread, but I can no longer allow myself to get involved where you
> offer only gross generalizations with no attempt toward supplying pertinent
> and applicable details of any kind.
>

[Gone Angling]

>Try Wal-Mart, they are in the fishing section that you claim to frequent.
>

Mr. Coger they don't seem have one at my Walmart. I was there this morning. I
was thinking of one of the boat dealers may have one at one of the upcoming
fishing shows. I would prefer to buy one at one of the volume outlets as you
suggest. walmart.

[Charles B. Summers]

No way you could have had three girls AND a sheep! Quit yer braggin'. No-one
could be that lucky...


"Fritz Nordengren" > wrote in message
news:grNrb.162490$e01.576463@attbi_s02...
If you didn't have those
> pictures of me with those three girls and the sheep, I never would have
> covered for you.

[Chuck Coger]

I was at my Wal-Mart and K-Mart this morning and they both have them. Seats
and Poles, they are in the boating section. If not pickup the yellow pages
or go to www.switchbard.com and do a search for boat dealers in your area.
Or order them from BPS, Cabelas, E-Bay, Boaters Worlds or Overtons. The are
100's of places that have them.

---
Chuck Coger
http://www.fishin-pro.com


"Gone Angling" > wrote in message
...
> >Try Wal-Mart, they are in the fishing section that you claim to frequent.
> >
>
> Mr. Coger they don't seem have one at my Walmart. I was there this
morning. I
> was thinking of one of the boat dealers may have one at one of the
upcoming
> fishing shows. I would prefer to buy one at one of the volume outlets as
you
> suggest. walmart.
>
>
>

[Henry Hefner]

Maybe he meant he had 3 girls BY the sheep? .... ;)

Charles B. Summers wrote:
> No way you could have had three girls AND a sheep! Quit yer braggin'. No-one
> could be that lucky...
>
>
> "Fritz Nordengren" > wrote in message
> news:grNrb.162490$e01.576463@attbi_s02...
> If you didn't have those
>
>>pictures of me with those three girls and the sheep, I never would have
>>covered for you.
>
>
>

[Bob Rickard]

Fritz, if I had a better supplier for my crayons I would have never needed
your help in Vegas. You just try calculating the square root of a green lump
sometime! Besides, if you shared the sheep like I asked, I wouldn't have had
to report you to Al, who promptly came & took it back home with him. So
there now!
--
Bob Rickard
www.secretweaponlures.com
--------------------------<=x O')))><


"Fritz Nordengren" > wrote in message
news:grNrb.162490$e01.576463@attbi_s02...
> Now Rickard, I've been polite to you on this group to gave you a second
> chance, but surely you remember that incident in the late 80's at Los
> Alamos where the entire physics staff discredited your so called
> "theories" on string cheese as a fossil fuel.
>
> Rickard, when will you learn? there are those of us who are real
> scientists and not some bass fishing hobbiest who thinks that just
> because he knows E=Mc2 can tell those of us in the know how to
> differentiate a calculated mass.
>
> And also, don't forget, who was there to save your butt in Las Vegas
> during the 1991 International Conference on Refractive Indulgence. If
> Ihadn't spent three hours correcting your formulas, you would have been
> driven from the scientific community for life. If you didn't have those
> pictures of me with those three girls and the sheep, I never would have
> covered for you.
>
> So lets just let it pass and agree to disagree.
>
> <g>
>
>
>
> Bob Rickard wrote:
> > Fritz, I would contribute my usual definitive, leading edge
contributions to
> > this thread, but I can no longer allow myself to get involved where you
> > offer only gross generalizations with no attempt toward supplying
pertinent
> > and applicable details of any kind.
> >
>